Problems of Remainder Theory can be solved by pattern for 1-9 digit within 30 seconds.

2. Always try to co-relate power of numerator with denominator
Such that
𝑵 𝑷
𝒅
𝑹 = 𝟏 / 𝟎
EX
(𝟓 𝟕𝟐𝟑)
𝟑𝟏
=
𝟓 𝟑 𝟏𝟒𝟏
𝟑𝟏 𝒓 =1… … ∵
𝟏𝟐𝟓
𝟑𝟏 𝒓 = 𝟏

3. Sometime we need to manipulate to achieve that by using a
law of index
𝒂 𝒃
∗ 𝒂 𝒄
= 𝒂 𝒃+𝒄
E.g.
𝟓 𝟖𝟗
𝟑𝟏
=
𝟓 𝟖𝟕∗𝟓 𝟐
𝟑𝟏
=
(𝟓 𝟑)^𝟐𝟗 ∗𝟐𝟓
𝟑𝟏 𝐫
= 𝟏 ∗ 𝟐𝟓
= 𝟐𝟓

4. 𝒏𝒌 + 𝟏
𝒏 𝒓.
𝒐𝒅𝒅 /𝒆𝒗𝒆𝒏
= 𝟏
𝑬. 𝒈.
𝟑𝟗 𝟐𝟒𝟏
𝟏𝟗 𝒓 = 𝟏
… . ∵
𝟑𝟗
𝟏𝟗 𝒓 = 𝟏 𝒂𝒏𝒅 𝟏 𝟐𝟒𝟏 = 𝟏

5. 𝒏𝒌 − 𝟏
𝒏 𝒓
.
𝒐𝒅𝒅
= 𝒏 − 𝟏
E.g.
𝟐𝟒𝟗
𝟐𝟓 𝒓.
𝟏𝟑𝟕
=
𝟐𝟓𝟎−𝟏
𝟐𝟓 𝒓.
𝟏𝟑𝟕
=
−𝟏 𝟑𝟕
𝟐𝟓 𝒓.
.
=
−𝟏
𝟐𝟓 𝒓.
.
= 𝟐𝟒

6. 𝒏𝒌 − 𝟏
𝒏 𝒓
.
𝒆𝒗𝒆𝒏
= 𝟏
𝟏𝟐𝟎 𝟔𝟒
𝟏𝟕 𝒓.
𝟏𝟑𝟕
=
𝟏𝟏𝟗 + 𝟏
𝟏𝟕 𝒓.
𝟔𝟒
= 𝟎 + 𝟏 𝟔𝟒 .
= 𝟏

7. Method Illustration
If a given number NP is divided by
x then remainder would be (say R)
What would be remainder when
1610 is divided by 7?
𝑵 𝒑
𝒙 r =
𝒏𝒙+𝑨) 𝒑
𝒙
or
𝒏𝒙−𝑨) 𝒑
𝒙
Iterate the same to get following
expression
𝑵 𝒑
𝒙 r =
𝒏𝒙+𝑨) 𝒑
𝒙
or
𝒏𝒙−𝑨) 𝒑
𝒙
=
𝟏 𝒑
𝒙
𝒐𝒓 = −
𝟏 𝒑
𝒙
𝟏𝟔 𝟏𝟎
𝟕 r =
𝟐∗𝟕+𝟐 𝟏𝟎
𝟕
=
𝟐 𝟏𝟎
𝟕
=
(𝟐 𝟑) 𝟑
𝟕
∗ 𝟐 =
𝟖 𝟑
𝟕
∗ 𝟐
=
𝟕 + 𝟏 𝟑
𝟕
∗ 𝟐 =
𝟏 𝟑
𝟕
∗ 𝟐
R=7

8. Method Illustration
If p is even then remainder would
be same
What would be remainder when
213 is divided by 7?
(𝟐 𝟑)∗𝟐.
𝟕
=
𝟖 𝟒∗𝟐
𝟕
=
𝟕+𝟏 𝟒∗𝟐
𝟕
=
𝟏 𝟒 ∗ 𝟐
𝟕
∴ 𝑹 = 𝟐

9. Method Illustration
If p is odd then remainder would
be R=X – A
What would be remainder when
395 is divided by 10?
(𝟑 𝟐)^𝟒𝟕∗𝟑.
𝟏𝟎
=
𝟗 𝟒𝟕∗𝟑
𝟏𝟎
=
𝟏𝟎−𝟏 𝟒𝟕∗𝟑
𝟏𝟎
=
−𝟏 𝟒𝟕∗𝟑
𝟏𝟎
∴ 𝑹 = 𝟏𝟎 − 𝟐 ∗
𝟑
𝟏𝟎
= 𝟗 ∗
𝟑
𝟏𝟎
=
𝟐𝟕
𝟏𝟎
𝒐𝒓 𝑹 = 𝟕

10. Method Illustration
If the sum of N1+N2 is divided by X then
remainder would be ( say R)
What would be remainder when
10082+10085 is divided by 13?
(𝑵𝟏 + 𝑵𝟐)
𝒙 𝒓 =
𝑵𝟏
𝒙 𝒓 +
𝑵𝟐
𝒙 𝒓
=
𝑨 + 𝑩
𝒙 𝒓
(𝟏𝟎𝟎𝟖𝟐 + 𝟏𝟎𝟎𝟖𝟓)
𝟏𝟑 𝒓
=
𝟏𝟎𝟎𝟖𝟐
𝟏𝟑 𝒓 +
𝟏𝟎𝟎𝟖𝟓
𝟏𝟑 𝒓
=
𝟕+𝟏𝟎
𝟏𝟑 𝒓
𝑹 = 𝟒

11. 1
11 12 13 14 15 16 17
1 1 1 1 1 1 1
The number 1 is representing itself in each unit place of next number
2
21 22 23 24 25 26 27
2 4 8 16 32 64 128
The unit digit follows a pattern of 2,4,8,6
3
31 32 33 34 35 36 37
3 9 27 81 243 729 2187
The unit digit follows a pattern of 3,9,7,1

12. 4
41 42 43 44 45 46 47
4 16 64 256 1024 4096 16384
The unit digit follows a pattern of 4,6
5
51 52 53 54 55 56 57
5 25 125 625 3125 15625 78125
The unit digit follows a pattern of 5
6
61 62 63 64 65 66 67
6 36 216 1296 7776 46656 279936
The unit digit follows a pattern of 6,

13. 7
71 72 73 74 75 76 77
7 49 343 2401 16807 117649 823543
The unit digit follows a pattern of 7,9,3,1
8
81 82 83 84 85 86 87
8 64 512 4096 32768 262144 2097152
The unit digit follows a pattern of 8,4,2,6
9
91 92 93 94 95 96 97
9 81 729 6561 59049 531441 4782969
The unit digit follows a pattern of 9,1

14. 1. Find the Last digit of (32^32)^32.
2. Find the remainder when 7^7 is divided by 4.
3. Find the remainder when 11^8 is divided by 7.
4. Find the remainder when 923^888+235^222 is divided by 4.
5. Find the remainder when 123^321 is divided by 5.

15. 1. Find the Last digit of (32^32)^32.
Solution:
Step 1: The last digit of (32^32)^32 is same as (2^32)^32
Step 2: But (2^32)^32=2^(32×32×32×….×32 times)
= (2^32)^32=2^4×8×(31×32×…..32 times)
= (2^32)^32=2^4n,
Where n=8×(31×32×…..32 times)
Step 3: Again 2^4n=(16)^n = unit digit is 6,for every n€N
Step 4: Hence, the required unit digit = 6.

16. 2. Find the remainder when 7^7 is divided by 4.
Solution:
Step 1: 7^7/4=(7×7×7× 7×7×7×7)/4
Step 2: (3×3×3× 3×3×3×3)/4
Step 3: (9×9×9×3)/4
Step 4: (1×1×1×3)/4
Step 5: 3/4
Step 6: Thus the required remainder is 3.

17. 3. Find the remainder when 11^8 is divided by 7.
Solution:
Step 1: 11^8/7=(11×11×11×11×11×11×11×11)/7
Step 2: (4×4×4×4×4×4×4×4)/7
Step 3: 16×16×16×16/7
Step 4: 2×2×2×2/7
Step 5: 16/7
Step 6: Thus the required remainder is 2.

18. 4. Find the remainder when 923^888+235^222 is divided by 4.
Solution:
Step 1: Rem (923^888+235^222)/4
Step 2: rem.( 3^888+3^222)/4
Step 3: rem.(1+1)/4=2/4.
Step 4: Thus the remainder is 2.

19. 5. Find the remainder when 123^321 is divided by 5.
Solution:
Step 1: The remainder when 123 is divided by 5 is 3
Step 2: So, the remainder when 123^321 is divided by 5 is same as when
3^321 is divided by 5.
Step 3: Now, the remainder when 3^1 is divided by 5 is 3.
the remainder when 3^2 is divided by 5 is 4.
the remainder when 3^3 is divided by 5 is 2.
the remainder when 3^4 is divided by 5 is 1.
the remainder when 3^5 is divided by 5 is 3.
So the cycle period is 4 since at 3^4, we get the remainder 1
(After which the cycle starts repeating)

20. Step 4: Thus the remainder when 3^321 is divided by 5 is 3 since we get
the remainder 1 when 321 is divided by 4(the cyclic period)
Step 5: Or rem=(123^321)/5=rem.(3^321)/5=rem.(3^320×3^1)/5
= rem.(3^4)^80×3^1/5=rem.3^1/5
=remainder is 3.